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Thread: Series dealing with exponationals

  1. #1
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    Series dealing with exponationals

    I don't see how

    1. $\displaystyle \sum_{n=1}^\infty \frac{1}{e^n} + \frac{1}{n(n+1)}$ = $\displaystyle \frac{e}{e-1}$

    2. $\displaystyle \sum_{n=1}^\infty e^\frac{1}{n} - e^\frac{1}{n+1}$ = $\displaystyle e-1$

    In the first case I know that as $\displaystyle n\rightarrow\infty$ that $\displaystyle \frac{1}{e^n}$ approaches 0 and therefore it converges, and I know that $\displaystyle \frac{1}{n(n+1)} = 1$ (from a previous problem). But how do you know what $\displaystyle \frac{1}{e^n}$ sums up to?.

    And in the second case I have no clue about approaching the problem, but I would think that it's the same type of approach as the first.
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    \sum\limits_{n = 1}^\infty {\left( {\frac{1}
    {e}} \right)^n } $

    is just an infinite geometric series, and we know that it converges since 1/e < 1 thus:


    $\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{1}
    {e}} \right)^n } = \frac{{a_1 }}
    {{1 - q}} = \frac{{\frac{1}
    {e}}}
    {{1 - \frac{1}
    {e}}} = \frac{1}
    {{e - 1}}$

    the second term can be rewritten using partial fraction decomposition as:

    $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{1}
    {{n\left( {n + 1} \right)}} = } \sum\limits_{n = 1}^\infty {\frac{1}
    {n}} - \frac{1}
    {{n + 1}}$

    $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{1}
    {n}} - \frac{1}
    {{n + 1}} = \left( {1 - \frac{1}
    {2}} \right) + \left( {\frac{1}
    {2} - \frac{1}
    {3}} \right) + \left( {\frac{1}
    {3} - \frac{1}
    {4}} \right) + \cdots = 1
    $

    since both individual terms converge to 0.
    so:

    $\displaystyle
    \sum\limits_{n = 1}^\infty {\frac{1}
    {{e^n }}} + \sum\limits_{n = 1}^\infty {\frac{1}
    {{n\left( {n + 1} \right)}} = \frac{1}
    {{e - 1}} + 1 = \frac{e}
    {{e - 1}}}
    $

    $\displaystyle
    \sum\limits_{n = 1}^\infty {e^{\frac{1}
    {n}} - e^{\frac{1}
    {{n + 1}}} = \left( {e - e^{\frac{1}
    {2}} } \right) + \left( {e^{\frac{1}
    {2}} - e^{\frac{1}
    {3}} } \right) + \cdots }
    $

    now let's look at the sum of the first N elements and then take the limit as N approaches infinity:

    $\displaystyle
    \mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {e^{\frac{1}
    {n}} - e^{\frac{1}
    {{n + 1}}} = \left( {e - e^{\frac{1}
    {2}} } \right) + \left( {e^{\frac{1}
    {2}} - e^{\frac{1}
    {3}} } \right) + \cdots }
    $$\displaystyle
    + \left( {e^{\frac{1}
    {{N - 1}}} - e^{\frac{1}
    {N}} } \right) + \left( {e^{\frac{1}
    {N}} - e^{\frac{1}
    {{N + 1}}} } \right) = e - 1
    $
    Last edited by Peritus; Feb 17th 2008 at 11:49 PM.
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